A body of mass m is released from a height equal to the radius R of earth. The velocity with which it will strike earth’s surface is

Question

A body of mass m is released from a height equal to the radius R of earth. The velocity with which it will strike earth’s surface is (A) √2gR (B) √gR (C) √2mgR (D) √mgR

Tardigrade Admin · Accepted Answer

Since, the initial velocity of body is zero, its total energy is

E = \frac{- G m M}{r}

...(i)
where,

m

is the mass of body,

M

is the mass of the Earth and

r

its distance from the centre of the Earth. When the body reaches the Earth, let its, velocity be

v

and its distance from the centre of the Earth equals the Earths radius

R

. Therefore, the energy is

E = \frac{1}{2} m v^{2} - \frac{GM m}{R}

...(ii)
Equating Eqs. (i) and (ii), we get

v^{2} = 2 GM (\frac{1}{R} - \frac{1}{r})

Also,

g = \frac{GM}{R ^{2}}

. Therefore,

GM = g R^{2}

.
Using this in the above equation, we get

V = R [2 g (\frac{1}{R} - \frac{1}{r})]^{1/2}

Now,

r = 2 R

(given). Therefore,

V = R [2 g (\frac{1}{R} - \frac{1}{2 R})]^{1/2}

\Rightarrow

Velocity of a body which strike the Earth's surface,

v = g R

Q. A body of mass $m$ is released from a height equal to the radius $R$ of earth. The velocity with which it will strike earth’s surface is

$2 g R$

$g R$

$2 m g R$

$m g R$

$m g R$

Q. A body of mass m is released from a height equal to the radius R of earth. The velocity with which it will strike earth’s surface is

2gR​

gR​

2mgR​

mgR​

mgR​

Q. A body of mass $m$ is released from a height equal to the radius $R$ of earth. The velocity with which it will strike earth’s surface is

$2 g R$

$g R$

$2 m g R$

$m g R$

$m g R$